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Despite its formula, boric acid, H3BO3, is a monoprotic acid. A 25.00 mL sample of this...

Despite its formula, boric acid, H3BO3, is a monoprotic acid. A 25.00 mL sample of this acid required 15.66 mL of 0.1724 M NaOH to reach the equivalence point where the pH= 11.13. Determine the original concentration of boric acid and the equilibrium constant, Ka.

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Answer #1

millimoles of NaOH = 15.66 x 0.1724 = 2.70

at equivalence point .

millimoles of acid = millimoles of base

M x 25 = 2.70

M = 0.1080

concentration of H3BO3 = 0.1080 M .

H3BO3 +   NaOH   ------------->

at equivalence point salt only remains. it is the salt of strong base and weak acid.

salt concentration = C = 2.70 / (15.66 + 25) = 0.06640 M

pH = 7 + 1/2 (pKa + log C)

11.13 = 7 + 1/2 (pKa + log 0.0664)

pKa = 9.438

Ka = 10^-pKa = 10^-9.438

Ka of boric acid = 3.649 x 10^-10

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